What is the degree of the differential eqaution : $$\dfrac{d^3y}{dx^3}-6(\dfrac{dy}{dx})^2-4y=0$$

What are the order and degree, respectively, of the differential equation
$${ \left( \cfrac { { d }^{ 3 }y }{ d{ x }^{ 3 } }  \right)  }^{ 2 }={ y }^{ 4 }+{ \left( \cfrac { dy }{ dx }  \right)  }^{ 5 }\quad $$?

Solution of the differential equation
$$\tan { y } .\sec ^{ 2 }{ x } dx+\tan { x } .\sec ^{ 2 }{ y } dy=0$$ is

The order of differential equation corresponding to $$y={ c }_{ 1 }\cos { 2x } +{ c }_{ 2 }\cos ^{ 2 }{ x } +{ c }_{ 3 }\sin ^{ 2 }{ x } +{ c }_{ 4 }$$

The solution of $$\sec ^{2}y\dfrac {dy}{dx}+2x\tan y=x^{3}$$ is

The solution of $$\dfrac { dx }{ dy } -\dfrac { 2 }{ 3 } xy={ x }^{ 4 }{ y }^{ 3 }$$ is

Solution of $${y^2}dx + ({x^2} - xy + {y^2})dy = 0$$ 

If tangent at point p, with parameter t, on the curve $$x=4t^2+3$$,$$y=8t^3-1,t \epsilon R,$$ meets the curve again at point Q, then the coordinates of Q are:- 

The solution of $$\dfrac{dx}{dy} + \dfrac{x}{y} = x^2$$  is 

If $$y = cos^{-1} x$$ then $$(1 - x^2)y'' -xy'=?$$ where $$y'= \dfrac{dy}{dx}$$